Problem: Estimating $\cos(0.95)$ using a Maclaurin polynomial, what is the least degree of the polynomial that assures an error smaller than $0.001$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $3$ (Choice B) B $4$ (Choice C) C $5$ (Choice D) D $6$
Answer: We will use the Lagrange error bound. Let's assume the polynomial's degree is $n$. The absolute value of the $(n+1)^{\text{th}}$ derivative of $\cos(x)$ is bounded by $1$. [Why?] The Lagrange bound for the error assures that: $|R_n(0.95)|\leq \dfrac{1}{(n+1)!}0.95^{n+1}$ Solving $\dfrac{0.95^{n+1}}{(n+1)!}<0.001$ using trial and error, we find that $n\geq 6$. In conclusion, the least degree of the polynomial that assures our error bound is $6$.